Page 65 - Advanced Linear Algebra

Vector Spaces 49

Note that the first set of vectors still spans and the second set is still linearly
=
independent.
Now we repeat the process, moving from the second list to the first list
#

#Á #Á Á Ã Á Â #Á Ã Á #

As before, the vectors in the first list are linearly dependent, since they spanned
#
= # before the inclusion of . However, since the 's are linearly independent,

any nontrivial linear combination of the vectors in the first list that equals
must involve at least one of the 's. Hence, we may remove that vector, which

again by reindexing if necessary may be taken to be and still have a spanning

set
#Á #Á Á Ã Á Â #Á Ã Á #

Once again, the first set of vectors spans and the second set is still linearly
=
independent.
Now, if  , then this process will eventually exhaust the 's and lead to the

list
#Á #Á Ã Á # Â # b Á Ã Á #

=

where #Á #Á Ã Á # span , which is clearly not possible since # is not in the

. Hence,  .
span of #Á #Á Ã Á #

Corollary 1.11 If has a finite spanning set, then any two bases of have the
=
=
same size.
Now let us prove the analogue of Corollary 1.11 for arbitrary vector spaces.
Theorem 1.12 If is a vector space, then any two bases for have the same

=
=
cardinality.
Proof. We may assume that all bases for are infinite sets, for if any basis is
=
finite, then has a finite spanning set and so Corollary 1.11 applies.
=
9
=
=
8
Let ~¸ “ 0¹ be a basis for and let be another basis for . Then any

vector  9 can be written as a finite linear combination of the vectors in ,
8
where all of the coefficients are nonzero, say
~

<
But because is a basis, we must have
9
<~ 0

9